There are N children standing in a line. Each child is assigned a rating value.
You are giving candies to these children subjected to the following requirements:
- Each child must have at least one candy.
- Children with a higher rating get more candies than their neighbors.
What is the minimum candies you must give?
Example 1:
Input: [1,0,2] Output: 5 Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
Example 2:
Input: [1,2,2] Output: 4 Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively. The third child gets 1 candy because it satisfies the above two conditions.
Approach #1: C++. [Using two array]
class Solution { public: int candy(vector<int>& ratings) { int size = ratings.size(); int ans = 0; vector<int> ltor(size, 1); vector<int> rtol(size, 1); for (int i = 1; i < size; ++i) { int j = size - i - 1; if (ratings[i] > ratings[i-1]) ltor[i] = ltor[i-1] + 1; if (ratings[j] > ratings[j+1]) rtol[j] = rtol[j+1] + 1; } for (int i = 0; i < size; ++i) ans += max(ltor[i], rtol[i]); return ans; } };
Approach #2: Java. [Using one array]
public class Solution { public int candy(int[] ratings) { int[] candies = new int[ratings.length]; Arrays.fill(candies, 1); for (int i = 1; i < ratings.length; i++) { if (ratings[i] > ratings[i - 1]) { candies[i] = candies[i - 1] + 1; } } int sum = candies[ratings.length - 1]; for (int i = ratings.length - 2; i >= 0; i--) { if (ratings[i] > ratings[i + 1]) { candies[i] = Math.max(candies[i], candies[i + 1] + 1); } sum += candies[i]; } return sum; } }
Approach #3: Java. [Single Pass Approach with Constant Space]
class Solution { public int count(int n) { return (n * (n + 1)) / 2; } public int candy(int[] ratings) { if (ratings.length <= 1) return ratings.length; int candies = 0; int up = 0; int down = 0; int old_slope = 0; for (int i = 1; i < ratings.length; ++i) { int new_slope = ratings[i] > ratings[i-1] ? 1 : ratings[i] < ratings[i-1] ? -1 : 0; if (old_slope > 0 && new_slope == 0 || old_slope < 0 && new_slope >= 0) { candies += count(up) + count(down) + Math.max(up, down); up = 0; down = 0; } if (new_slope > 0) up++; if (new_slope < 0) down++; if (new_slope == 0) candies++; old_slope = new_slope; } candies += count(up) + count(down) + Math.max(up, down) + 1; return candies; } }
reference:
https://leetcode.com/problems/candy/solution/